INTEGRAL FUNCTIONS OF ORDER ONE 1. Jensen's formula Write ...

INTEGRAL FUNCTIONS OF ORDER ONE
PART III PRIME NUMBERS, MICHAELMAS 2004
1. Jensen’s formula
Write CR for the circular contour |z| = R and BR for the domain |z| < R.
Theorem 1.1 (Jensen’s formula). Let R, ɛ > 0. Suppose that f is analytic on BR+ɛ
that f(z) �= 0 for R � |z| < R + ɛ and for z = 0, and that f has zeros z1, . . . , zn in BR
(counted with multiplicity). Then
1
2π
� 2π
0
log |f(Re iθ )| dθ = log |f(0)| + log
Rn . (1.1)
|z1| . . . |zn|
Proof. Observe that if the identity is true for functions f1 and f2 then it is also true for
f1f2. Write
R(z − zj)
gj(z) =
R2 − zjz
and define a meromorphic function F by
f(z) = Cg1(z) . . . gn(z)F (z),
where C is chosen so that F (0) = 1. If ɛ is chosen so small that the poles z = R 2 /zj lie
outside of BR+ɛ then F has no zeros in BR+ɛ. Jensen’s formula being manifestly true
for constant functions, it suffices to check it for F and for the functions gj.
Now we may define a single-valued, analytic logarithm of F in BR+ɛ by the formula
�
F
log F (z) :=
′ (w)
F (w) dw.
[0→z]
Since F (0) = 1 the function z−1 log F (z) is also analytic in BR+ɛ. Hence by Cauchy’s
theorem we have �
log F (z)
dz = 0.
z
Parametrising CR by z = Re iθ , 0 � θ < 2π we get
Taking real parts gives
CR
� 2π
0
� 2π
0
log F (Re iθ ) dθ = 0.
log |F (Re iθ )| dθ = 0.
This is one side of (1.1); the other side is clearly zero. Thus we have verified the formula
for F .
1

2 PART III PRIME NUMBERS, MICHAELMAS 2004
Turning our attention to the functions gj, note that if |z| = R then
�
�
|gj(z)| = �
z(z − zj)
� R2 �
�
�
− zjz � = 1.
Thus the left-hand side of (1.1) equals 0. As for the right-hand side, note that |gj(0)| =
|z1|/R. Thus the right-hand side is zero as well.
Suppose that f is an entire function. Write n(r) for the number of zeros of f in Br.
Corollary 1.2. Let f be an entire function with f(0) �= 0. Then
1
2π
� 2π
0
log |f(Re iθ )| dθ − log |f(0)| =
Proof. By Jensen’s theorem it suffices to check that
log
R n
|z1| . . . |zn| =
� R
0
� R
0
r −1 n(r) dr.
r −1 n(r) dr. (1.2)
To do this, list the zeros in BR in order of magnitude, thus |zj| = rj and r1 � r2 �
. . . � rn. For notational convenience set r0 := 0 and rn+1 := R. We have
� R
r −1 n�
� rj+1
n(r) dr = r −1 n�
n(r) = j log(rj+1/rj) = log(R n /r1 . . . rn),
0
as required.
j=0
rj
j=0
2. Integral functions of order one
An integral function of order one is an entire function f : C → C satisfying the bound
|f(z)| = Oɛ(exp(|z| 1+ɛ ))
for all ɛ > 0. We will encounter several such functions in this course. Two examples are
1/Γ(z) and the “super-completed” zeta function ξ(z) which we will encounter in [PN3].
The next lemma allows us to discern the structure of integral functions of order one
which have no zeros. We will use it to prove Theorem 2.2, and it is to fit that need that
we operate under what may seem at this point like unnecessarily weak hypotheses.
Lemma 2.1. Suppose that g is an entire function with no zeros which satisfies the
bound
|g(z)| = exp(O(|z| 3/2 ))
for all z ∈ �∞ j=1 CRj , where Rj → ∞ as j → ∞. Then g(z) = eAz+B for some constants
A, B.
Proof. This is a simple variant of a well-known argument used to prove Lioville’s theorem.
First of all note that g has an analytic logarithm h(z) := log g(z), which is subject
to the bound |h(w)| = O(|w| 3/2 ) whenever |w| = Rj for some j. Fix z ∈ C. By Cauchy’s
derivative formula we have
h ′′ (z) = 1
�
iπ CR
h(w) dw
(w − z) 3

INTEGRAL FUNCTIONS OF ORDER ONE 3
provided that |z| < R, which implies that
|h ′′ (z)| � 1
�
|h(w)||w − z|
π
−3 dw. (2.1)
CR
Pick j large enough that z ∈ BRj/2, and apply (2.1) with R = Rj. We have the crude
estimates |w − z| ≫ Rj for w ∈ CRj and, by assumption, |h(w)| ≪ R3/2 j . Also, the
length of the path of integration is 2πRj. Thus estimating (2.1) trivially yields
|h ′′ (z)| ≪ R −1/2
j .
Letting j → ∞ we are forced to conclude that h ′′ (z) = 0, which means that h(z) =
Az + B for some constants A, B.
We turn now to the main result of this set of notes, a structure theorem for integral
functions of order one.
Theorem 2.2. Suppose that f is an integral function of order 1 and that X > 1. Then
(i)
�
|ρ|�X
|ρ| −1−ɛ ≪ɛ X −ɛ/2
for any ɛ > 0, where the sum ranges over all zeros ρ of f, counted with multiplicity.
(ii) The sum
�
|ρ| −1−ɛ
converges, for any ɛ > 0.
(iii) We have the product representation
for some constants A, B.
f(z) = z r e
ρ�=0
Az+B �
ρ�=0
(1 − z/ρ)e z/ρ
Proof. (i) If f has a zero of order r at 0 then we may divide through by zr . The resulting
function is still entire and has order 1, but now f(0) �= 0. By Corollary 1.2 we have
� R
0
r −1 n(r) dr = 1
2π
� 2π
0
log |f(Re iθ )| dθ − log |f(0)| ≪ɛ R 1+ɛ
for all ɛ > 0. Writing SR for the number of zeros of f in the annulus R � |z| < 2R, we
therefore have
SR � 1
� 3R
n(r) dr ≪ɛ R
R 2R
1+ɛ/2 .
Thus
�
|ρ| −1−ɛ �
∞�
2 −(1+ɛ)j S2j ≪ɛ
�
2 −ɛj/2 ≪ɛ X −ɛ/2 ,
as required.
|ρ|�X
j:2 j+1 �X
j:2 j+1 �X
(ii) This follows immediately from (i) and a compactness argument (to deal with potential
zeros having |ρ| < 1).

4 PART III PRIME NUMBERS, MICHAELMAS 2004
(iii) A consequence of (ii), with ɛ = 1, is that
�
|ρ| −2 < ∞.
This implies that the function
ρ�=0
F (z) := �
(1 − z/ρ)e z/ρ
ρ�=0
is entire and has zeros precisely at the zeros ρ of f (we will not prove this fact). Thus
g(z) := f(z)/z r F (z) is an entire function with no zeros. We wish to prove that in fact
g(z) = e Az+B , for some constants A, B, and to do this we will apply Lemma 2.1. That
lemma requires us to give an upper bound for g, but only on a series of circular contours
CRj ; it clearly makes sense to choose these contours so as to avoid the zeros ρ, and this
is what we do.
For each j � 1, we choose a value of Rj ∈ [2j , 2j+1 ). The number of zeros in the annulus
2j � |z| < 2j+1 is Oɛ(2j(1+ɛ) ), and so (choosing ɛ = 1) there is some R ∈ [2j , 2j+1 ) such
that
|z − ρ| ≫ R −1
(2.2)
for all z ∈ CR and all zeros ρ. Set Rj := R, and throughout what follows suppose that
|z| = R. To get a lower bound on
E := �
(1 − z/ρ)e z/ρ ,
divide the product into three parts
where
and
E1 :=
E2 :=
ρ�=0
E = E1E2E3,
�
0

INTEGRAL FUNCTIONS OF ORDER ONE 5
the implied constant will depend on the location of any zeros ρ close to 0. Putting all
of this together gives
|E1| ≫ e −2R3/2
. (2.3)
Turning to E2, we have for R/10 < |ρ| � 10R the bound
|1 − z/ρ| � cR −2 ,
this following immediately from (2.2). Furthermore |e z/ρ | � c ′ > 0, and the total number
of ρ under consideration is ≪ɛ R 1+ɛ for all ɛ > 0. Thus we have
|E2| ≫ (cc ′ R −2 ) CɛR1+ɛ
≫ e −R3/2
. (2.4)
To bound E3, observe that by a crude Taylor series expansion one has
for |w| < 1/10. Thus
|E3| > �
|(1 − w)e w | > e −10|w|2
ρ:|ρ|>10R
e −10R2 |ρ| −2
> e −CR3/2
for some C, this last inequality being a consequence of (i).
Putting all this together gives
|E| = |E1||E2||E3| ≫ e −C′ R 3/2
.
It follows immediately that for |z| = R we have
|g(z)| = exp(O(R 3/2 )).
Furthermore, we have proved this for some R = Rj in every dyadic interval [2 j , 2 j+1 ),
j � 0. Thus the hypotheses of Lemma 2.1 are satisfied and we do at last see that
g(z) = e Az+B for some constants A, B.